2 Answers
2

If the result holds for $a$, then it holds for $a+1$: we know that $a\in V_{a+1}=\mathcal{P}(V_a)$; we need to show that $a\cup\{a\}\in \mathcal{P}(V_{a+1})$. Since $a\in V_{a+1}$, then $a\subseteq V_a$. Since $a$ is an ordinal, if $b\in a$, then $b\subseteq a$, hence $b\subseteq V_a$; thus, $b\in\mathcal{P}(V_a) = V_{a+1}$. Thus, for every $b\in a$, $b\in V_{a+1}$, which shows that $a\subseteq V_{a+1}$. Therefore, since $a\in V_{a+1}$ and $a\subseteq V_{a+1}$, it follows that $a\cup\{a\}\subseteq V_{a+1}$, hence $a\cup\{a\}=a+1\in \mathcal{P}(V_{a+1})=V_{a+2}$, as desired.

Finally, assume that $b$ is a limit ordinal and for all $a\lt b$ we have $a\in V_{a+1}$. We want to show that $b\in V_{b+1}$.

If $a\lt b$, then $a\in V_{a+1}\subseteq V_b$; since $b$ is an ordinal, this implies that if $a\in b$, then $a\in V_b$. Hence, $b\subseteq V_b$, so $b\in\mathcal{P}(V_{b}) = V_{b+1}$, as desired.

Case $\alpha = \emptyset$: We trivially have that for all $a$ in $V_\emptyset = \emptyset$ that $a \subset V_\emptyset$.

Case $\alpha = \beta + 1$ is a successor ordinal:
Assume that for all $a \in V_\beta$ we have $a \subset V_\beta$ and let $b \in V_{\beta + 1} = P(V_\beta)$. Then for each $x$ in $b$ we have that $x \in V_\beta$. By assumption we have $x \subset V_\beta$, so for each $y \in x$ we have $y \in V_\beta$. Hence for each $y$ in $x$, $\{y \} \in P(V_\beta)$ and hence $\bigcup_{y \in x} \{y\} = x \in P(V_\beta) = V_{\beta + 1}$. So for each $x$ in $b$ we have $x \in V_{\beta + 1}$ which is the same as saying $b \subset V_{\beta + 1}$.